How do you determine the limit of #4/(x-5)^2# as x approaches 5?

1 Answer
Konstantinos Michailidis
Mar 24, 2016

The limit #lim_(x->5) 4/(x-5)^2=1/0# which is an undefined form.This means that the limit does not exist there.

Taking side-limits we have that #lim_(x->5^+) 4/(x-5)^2=+oo# (values of x from the right of point #x=5#) and #lim_(x->5^-) 4/(x-5)^2=+oo# (values of x from the left of point #x=5#)

Since #4/(x-5)^2->+oo# increases without bound from both sides of point #x=5# we say that

#lim_(x->5) 4/(x-5)^2=+oo#

Note that this is not saying that the limit exists, since
#oo# is not a real number.

The graph of #4/(x-5)^2# around #x=5# explains visually our result.

enter image source here

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